Indeterminate forms of Limits

Examples with detailed solutions and exercises that solves limits questions related to indeterminate forms such as: \[ \dfrac{\infty}{\infty}, \quad 0^0, \quad \infty^0, \quad 1^\infty, \quad \infty \cdot 0, \quad \infty - \infty \]

Theorem

A second version of L'Hopital's rule allows us to replace the limit problem \[ \dfrac{\infty}{\infty} \] with another simpler problem to solve. If \(\lim f(x) = \infty\) and \(\lim g(x) = \infty\) and if \(\lim \dfrac{f'(x)}{g'(x)} \) has a finite value \(L\), or is of the form \(\infty\) or \(-\infty\) then \[ \lim \dfrac{f(x)}{g(x)} = \lim \dfrac{f'(x)}{g'(x)} \] In the above \(\lim\) stands for \(\lim_{x \to a} f(x)\), \(\lim_{x \to a^+} f(x)\), \(\lim_{x \to a^-} f(x)\), \(\lim_{x \to \infty} f(x)\) or \(\lim_{x \to -\infty} f(x)\).

Examples with Solutions

Example 1

Find the limit \[ \lim_{x \to \infty} \dfrac{\ln x}{x} \]

Solution to Example 1:

Since \[ \lim_{x \to \infty} \ln x = \infty \] and \[ \lim_{x \to \infty} x = \infty \] we have the indeterminate form \[ \lim_{x \to \infty} \dfrac{\ln x}{x} = \dfrac{\infty}{\infty} \] The above L'Hopital's rule can be used to evaluate the given limit question as follows \[ \lim_{x \to \infty} \dfrac{\ln x}{x} = \lim_{x \to \infty} \dfrac{(\ln x)'}{(x)'} \] Evaluate the derivatives in the numerator and the denominator \[ = \lim_{x \to \infty} \dfrac{1/x}{1} \] Evaluate the limits in the numerator and denominator \[ \lim_{x \to \infty} (1/x) = 0 \text{ and } \lim_{x \to \infty} 1 = 1 \] We now evaluate the given limit \[ \lim_{x \to \infty} \dfrac{\ln x}{x} = \lim_{x \to \infty} \dfrac{0}{1} = 0 \]

Example 2

Find the limit: \[ \lim_{x\to \infty} x e^{-x} \]

Solution to Example 2:

Note that \[ \lim_{x\to \infty} x = \infty \] and \[ \lim_{x\to \infty} e^{-x} = 0 \] This is the indeterminate form \( \infty \cdot 0 \). The idea is to convert it into to the indeterminate form \( \dfrac{\infty}{\infty} \) and use L'Hopital's theorem. Note that \[ \lim_{x\to \infty} x e^{-x} = \lim_{x\to \infty} \dfrac{ x}{ e^x} = \dfrac{ \infty }{ \infty }\] We apply the above L'Hopital's theorem \[ = \lim_{x\to \infty} \dfrac{ (x)'}{ (e^x)'} = \lim_{x\to \infty} \dfrac{ 1 }{ e^x} = 0\]

Example 3

Find \[ \lim_{x\to \infty} ( 1 + 1/x)^x \]

Solution to Example 3:

Note that \( \lim_{x\to \infty} (1 + 1/x) = 1 \) and the above limit is given by \[ \lim_{x\to \infty} ( 1 + 1/x)^x = 1^{\infty}\] which is of the Indeterminate form \( 1^{\infty}\). If we let \( t = 1 / x \) the above limit may written as \[ \lim_{x\to \infty} ( 1 + 1/x)^x = \lim_{t\to 0} ( 1 + t)^{1/t} \] note that as \( x\to \infty \) , \( t\to 0 \)
Let \( y = ( 1 + t)^{1/t} \) and find the limit of \( \ln y \) as t approaches 0 \[ \ln y = \ln ( 1 + t)^{1/t} =(1 / t) \ln (1 + t) \] The advantage of using \( \ln y \) is that \[ \lim_{t\to 0} \ln y = \lim_{t\to 0} \dfrac {\ln (1 + t)}{t} = \dfrac{0}{0} \] and the limit has the indeterminate form 0 / 0 and the first L'Hopital's rule can be applied \[ \lim_{t\to 0} \ln y = \lim_{t\to 0} \dfrac {ln (1 + t)}{t} \] \[ = \lim_{t\to 0} \dfrac{(ln (1 + t))'}{(t)'} = \lim_{t\to 0} \dfrac{1/(1+t)}{1} = 1 \] Since the limit of \( \ln y = 1 \) the limit of \( y \) is \( e^1 = e \), hence \[ \lim_{x\to \infty} ( 1 + 1/x)^x = e \]

Example 4

Find the limit \[ \lim_{x\to 0^+} ( 1 / x - 1 / \sin x )\]

Solution to Example 4:

Note that \[ \lim_{x\to 0^+} ( 1 / x ) = + \infty \] and \[ \lim_{x\to 0^+} ( 1 / \sin x ) = + \infty \] This limit has the indeterminate form \( \infty - \infty \) and has to be converted to another form by combining \( 1 / x - 1 / sin x \) \[ \lim_{x\to 0^+} ( 1 / x - 1 / \sin x ) = \lim_{x\to 0^+} \dfrac{\sin x - x}{x \sin x} = \dfrac{0}{0} \] We now have the indeterminate form 0 / 0 and we can use the L'Hopital's theorem. \[ \lim_{x\to 0^+} \dfrac{\sin x - x}{x \sin x} = \lim_{x\to 0^+} \dfrac{(\sin x - x)'}{(x \sin x)'} = \lim_{x\to 0^+} \dfrac{\cos x - 1}{ \sin x + x \cos x} = \dfrac{0}{0} \] We have again the indeterminate form 0 / 0 and use the L'Hopital's theorem one more time. \[ = \lim_{x\to 0^+} \dfrac{(\cos x - 1)'}{ (\sin x + x \cos x)'} \] \[ = \lim_{x\to 0^+} \dfrac{-\sin x}{ \cos x + \cos x - x \sin x } = \dfrac{0}{2} = 0\]

Example 5

Find the limit \[ \lim_{x\to 0^+} x^x \]

Solution to Example 5:

We have the indeterminate form \( 0^0 \). Let \( y = x^x \) and \( \ln y = \ln (x^x) = x \ln x \). Let us now find the limit of \( ln y \) \[ \lim_{x\to 0^+} \ln y = \lim_{x\to 0^+} x \ln x = 0 \cdot \infty\] The above limit has the indeterminate form \( 0 \cdot \infty\). We have convert it as follows \[ \lim_{x\to 0^+} x \ln x = \lim_{x\to 0^+} \dfrac{\ln x}{1/x} = \dfrac{\infty}{\infty} \] It now has the indeterminate form \( \dfrac{\infty}{\infty} \) and we can use the L'Hopital's theorem \[ \lim_{x\to 0^+} \dfrac{\ln x}{1/x} = \lim_{x\to 0^+} \dfrac{(\ln x)'}{(1/x)'} \] \[ = \lim_{x\to 0^+} \dfrac{1/x}{-1/x^2} \] \[ = \lim_{x\to 0^+} - x = 0 \] The limit of \( ln y = 0 \) and the limit of \( y = x^x \) is equal to \[ \lim_{x\to 0^+} x^x = e^0 = 1 \]

Exercises

Find the limits

\( \quad \lim_{x\to \infty} (\ln x)^{1/x} \)
\( \quad \lim_{x\to \infty} (\ln x - \ln (1 + x)) \)
\( \quad \lim_{x\to \infty} \dfrac{x}{e^x} \)
\( \quad \lim_{x\to 0^+} x^{\sin x} \)